Mathematics Tricks (2nd Edition) C-O

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Type#5: 1/2α , 1/2β
Method: Multiply “b” term with 2 and “c” term with its square
and then interchange the coefficient of “a” and “c”.
Example: Let ax2+bx+c=0
Using the trick:
ax2+(2)b x+ c(c2)=0
c3x2+2bx+a=0
(Note: These tricks can be verified by usual method).

~Tricks for Solving Simultaneous equation:

Trick#1:

Put the points as given in all the options and satisfied the

given equations.

Trick#2:

Let we have two equations:

ax+by=c----- (i)

dx+ey=f------(ii)

For value of x: ( )− ()
( )− ()
=

For value of y:

Put the value of x in eq (i)....

Example:

Find the values of x and y by solving equation:

x+2y=10---- (i)

2x+3y=18----- (ii)

Using the Trick:

For x value: 2(18) − 3(10) 6
2(2) − 3(1) 1
= = = 6

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For y value:
Put x in (i):

6+2y=10
y=2
so x=6 and y=2. (Ans)

Judgment of Roots of Any Equation:-
If an equation and its roots are given in MCQ, then you can
simply check your answer by putting the roots in the
equation instead of solving the equation. The roots of the
equation satisfy the equation. e.g.

6 and -2 are the roots of given equation. If we put these roots in the
equation, it will satisfy the equation.

x=6: 2(6)²-8(6)-24 = 0
x=-2: 2(-2)²-8(-2)-24= 0 Both 6 and -2 satisfy the equation.

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Relation between Roots and Co-efficient:-

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Formation of the Equation by roots:
We can form equation by,

x²-Sx+P = 0

whereas, S = Sum of Roots
P= Product of Roots

 If an equation ax²+bx+c=0 is given and we have to find an
equation whose roots are n times the roots of given
equation, then equation can simply be found as;
ax²+n(b x)+n²c=0

 If the sum of the roots is given and we have to find the
sum of the roots raises to some power n.
Sum of roots of nth power = n (S)
Example: If the sum of the roots is 6 then find the sum of
the roots raises to power 3?
Using the trick:
Sum of roots of 3rd Power= 3(6)=18 (Answer)

 If the product of the roots is given and we have to find the
product of the root raises to some power n.
Product of roots of nth power= n2 (P)

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Example: If the sum of the roots is 8 then find the product

of the roots raises to power 3.

Using the trick:

Sum of roots of 3rd power= 32(8)=9(8)=72 (Answer)

The Quadratic equation when sum and product of roots raises to
some power n.
x2- n(S)x+n2P=0

Remainder Theorem:
If a polynomial f(x) of degree n≥1,where n is non-negative integer, is
divided by x-a till no x-term exists in the remainder then f(a) is the
remainder.
e.g. If we divide f(x)= x2+3x+7 by x+1 then remainder is
f(-1)= (-1)2+3(-1)+7=5

Factor Theorem:
The polynomial x-a is the factor of the polynomial f(x) if and if
f(a)=0
Nature Of Roots of the Quadratic Equation:
Nature of roots depends on the expression b2-4ac which is called
discriminant and denoted by D.
a) If D<0, then roots are imaginary
b) If D>0, then roots are real and distinct
c) If D=0, then roots are real and equal
d) Roots are rational iff D is a perfect square.
e) Roots are irrational iff D is positive but not a perfect square.
-Common Roots:
1) One Common Root:
If X is a common root of the equations:
a1 x2+b1x+c1=0 (1)
a2 x2+b2x+c2=0 (2)
then we have common root:

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X=

2) Both Common Roots:
If the equations (1) and (2) have both roots common then these
equations will be identical. Thus the required condition for both

root common is = =

Note:
1) To find the common root of two equations make the
coefficient of second degree terms in two equations equal and
subtract . The value of x so obtained is the required root.
2) If two quadratic equations with real coefficients have an
imaginary root common then both roots will be common and the

equations will be identical. The required condition is = =

3) If two quadratic equations have an irrational root common
then both roots will be common and the two equations will be

identical. The required condition is = =

Greatest and Least Value of a Quadratic Expression:
1) If a>0, then the quadratic expression y=ax2+bx+c has no

greatest value but it has least value. at x= .

2) If a<0, then the quadratic expression y=ax2+bx+c has no least

value but it has greatest value. . at x= .

Properties of Cube Root of Unity:
1) Each Complex cube root of unity is square of the other.

If √ = then If √ = and If √ = then

√=

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2) The sum of all the three cube root of unity is zero i.e
1+w+w2=0

3) The product if all the three cube root of unity is 1. i.e w3=1
4) For any n , wn is equivalent to one of the cube root of unity.
5) 1,w,w2 is a geometric sequence with common ratio “w”
6) {1,w,w2} is an Abelian Group under multiplication.
7) Each complex cube root of unity is square, square root,
reciprocal and conjugate of other.
8) wn+wn+1+wn+2=0 ,n
9) wn.wn+1 .wn+2=1 ,n
-Properties of Forth root of Unity:

1) Sum of all the four roots of unity is zero.
:. 1+(-1)+i+(-1)=0
2) Product of all the four roots of unity is -1
:. 1*(-1)*i*-i=1

3) The real fourth root of unity are additive inverses of each
other +1 and -1 are the real fourth root of unity and +1+(-
1)=0=(-1)+1
4) Both Complex imaginary fourth roots of unity are conjugate

,additive inverse and multiplicative inverse of each other.
Note: Number of roots of a polynomial is equal to degree of
polynomial.
-Quadratic inequalities Tricks:
1) (x-a)(x-b)>0  x<a or x>b for a<b
2) (x-a)(x-b)<0  a<x<b, for a<b
3) |x|<a  -a<x<a
4) |x|>a  x<-a or x>a

For Example: The solution set of |x-5|<9=?
So using the trick: -9<x-5<9 adding the 5 on thrice sides so that
-4<x<14 i.e (-4,14) (Answer)
For Example: The solution set of x2-5x+6<0=?

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First of all factorize it; (x-2)(x-3)<0
So by using the trick: x<2 or x<3 so ]2,3[ (Answer)
For Example: The Solution set of x2-1>0
As it is (x+1)(x-1)>0
So that -1<x<1 i.e (-1,1) (Answer)
For Example: The solution set of |x-1|>2
So using the Trick: x-1<-2 or x-1>2 i.e x<-1 ,x<3 (-1,3) ans
Note:
There are many types of question in which the quadratic
equation is not given but the method to solve is only by making
it a quadratic equation and then to solve them further. For these
questions, there are some tricks.

Tricks:

1) + + √ + ⋯ … … … . +∞=k+1

2) − − √ − ⋯ … … … . −∞=k then n=k(k+1)

3) + − √ + ⋯ … … … . ∞ = √

4) − + √ − ⋯ … … … . ∞ = √
5) If ax+bx=ak+bk ; a,b,x,k R then x=±k
6) If √Q1-√Q2=a-b ,a-b R such that Q1-Q2=a2-b2 then Q1=a2

For example: x= + + √ + ⋯ … … … . +∞ then x=?
Usual Method: squaring both sides.

x2=2+ + + √ + ⋯ … … … . +∞

x2=2+x => x2-x-2=0 => x=-1 and x=2
By trick: n= k(k+1) 2×1=k(k+1) so that x=2

For Example: x= + − √ + ⋯ … … … . ∞ Then Find the value of x?

By trick: x= √ so that x= ( ) =√

For Example: (4+√15)x+(4-√15)x =62, find the value of x?
Solution: As (4+√15)x+(4-√15)x =82+(-(√2)2),

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Here k=2 so that Using the trick; x=±k x=±2 (Answer)
For Example: √ − + − √ − + = , x=?
As − + − + − =3 such that 3=22-12
So Using the trick x2-5x+10=22  x2-5x+6=0 ,so x=2,3 (Answer)

Entry Test Typed MCQs:
1.The quadratic equation whose roots are 4 and -4 is given by :

A. x2+3x-28=0
B. x2-3x+28=0

C. x2-3x-28=0
D. x2+3x+28=0

Answer: Option C.

Explanation:

Let α =7 and β =-4.

Then, α+ β =3, α β = -28.

x2-(α+ β)x+ α β =0.

or x2-3x-28 = 0.

2.If some of the roots of a quadratic equation is 6 and the product of its roots is also
6, then the equation is
A. x2+6x-6=0
B. x2-6x+6=0
C. x2-6x-6=0
D. x2+6x+6=0

Answer: Option B.
Explanation:
α+ β =6 and α β = 6
The equation is :

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x2-(α+ β)x+ α β =0.
or x2-6x+6 = 0.
3. The quadratic equation with rational coefficients and having (2-√ 3 ), as one of its
roots is :

A. x2+4x+1=0
B. x2+4x-1=0
C. x2-4x-1=0
D. x2-4x+1=0

Answer: Option D.
Explanation:
Let α =(2-√ 3 )
Then β =(2+√ 3 )

α+ β =4 and α β =(4-3) = 1.
The equation is :
x2-(α+ β)x+ α β =0.
or x2-4x+1 = 0.

4.The quadratic equation with rational coefficients, one of whose roots is (3+2√ 3 ),
is:

A. x2+6x-3=0
B. x2-6x-3=0
C. x2+6x+3=0
D. x2-6x+3=0

Answer: Option B.
Explanation:

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Let α =(3+ 2√ 3 )
Then β =(3- 2√ 3 )

α+ β =6 and α β =(9-12) = -3.
The equation is :
x2-(α+ β)x+ α β =0.
or x2-6x-3 = 0.

5. The quadratic equation with rational coefficients and having (√ 2 -1) as one of its
roots is :

A. x2-2√ 2 x+1=0
B. x2-2√ 2 x-1=0
C. x2+2x-1=0
D. x2-2x+1=0

Answer: Option C.
Explanation:
Let α =(-1+ √ 2 )
Then β =(-1- √ 2 )

α+ β =-2 and α β = (1-2) =-1.
The equation is :
x2-(α+ β)x+ α β =0.
or x2+2x-1 = 0.

6.The quadratic equation with real coefficients and having (2+3i)as one of its roots,
is

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A. x2+4x+13=0
B. x2+4x-13=0
C. x2-4x-13=0
D. x2-4x+13=0

Answer: Option C.
Explanation:
Let α =(2+3i) and β =(2-3i).
Then, α+ β =4, α β = (4+9) = 13.
x2-(α+ β)x+ α β =0.
or x2-3x-28 = 0.
7. The quadratic equation with real coefficients and having (4+ √ -3 ) as one of its
roots, is :

A. x2-8x+19= 0
B. x2-8x-19= 0
C.x2+8x+19= 0
D. None of these

Answer: Option A.
Explanation:
Let α =(4+ √ 3i )
Then β =(4- √ 3i )

α+ β = 8 and α β = (16+3) =19.
The equation is :
x2-(α+ β)x+ α β =0.
or x2-8x+19 = 0.

8. If α, β are the roots of the equation x2-q(1+x)-r = 0, then the value of (1+α) (1+β) is

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A. (1-r)
B. (1+r)
C. (q-r)
D. (q+r)

Answer: Option A.
Explanation:
x2-qx-(q+r) = 0.
α+ β =q, α β = -(q+r).
(1+α)(1+β)
= 1+(α+ β)+α β
=1+q-(q+r)=(1-r).
9.If α, β are the roots of the equation x2-px+q = 0, then the value of (α2+β2) is

A. p2+2q
B. p2-2q
C. p(p2-3q)
D. p2-4q

Answer: Option B.
Explanation:
α+ β =p, α β = q
(α2+ β2) =(α+ β)2-2α β
=(p2-2q)
10.If the product of the roots of x2-3x+k is -2, then find the value of k?

A. -2
B. 8

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C. -8
D. 12

Answer: Option B.
Explanation:
Given equation is x2-3x+(k-10) =0.
∴ k-10 =-2
⇒ k =8.
11.If one root of 3x2-6kx+8k = 0 is 4, the other root is:

A. 2
B. -2
C. -4
D. 3

Answer: Option A.
Explanation:
x= 4 satisfies 3x2-6kx+8k =0.
∴ 3*16-6*k*4+8k = 0
⇒ 16k =48 i.e k =3.

∴ The equation is 3x2-18x+24 =0.
⇒ x2-6x+8 =0.

12.The quadratic equation whose roots are the reciprocals of the roots of the
equation ax2+bx+c = 0 is

A. cx2 +bx+c = 0
B. bx2 +cx+a = 0
C. ax2+bx2+c = 0
D. None of these

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Answer: Option A.
Self Explanatory.
18.If α, β be the irrational roots of ax2+bx+c = 0, where a, b, c are rational and a ≠ 0,
then

A. α = β
B. α,β = 1
C. α2+ β2
D. 1

Answer: Option D.
Explanation:
Clearly Option D is true.
13. The coefficient of x in the equation x2+px+q = 0 was taken as 17 in place of 13
and its roots were found to be -2 and -15. The roots of the original equation are:

A. 2, 15
B. 10, 3
C. -2, 15
D. -10, -3

Answer: Option D.
Explanation:
Let α,β be the roots of the original equation. then,
α+β = -13 and αβ =(-2)(-15) =30.
∴ original equation is
x2+13x+30 = 0.

⇒ x2+10x+3x+30 = 0.
⇒ x(x+10)+3(x+10) = 0.

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⇒ (x+10) (x+10) = 0.
⇒ x =-10 or x =-3.

14.The number of real solutions of x2-3|x|+2 = 0 is

A. 1
B. 2
C. 3
D. 4

Answer: Option D.
Explanation:
Case I: When x≥0
In this case, |x| =x.
So the equation is
x2-3x+2 = 0.
∴ (x-2)(x-1) =0.
∴ x =2 or x =1.

Case II: When x<0
In this case, |x| =-x.
So the equation is
x2+3x+2 = 0.
∴ (x+2)(x+1) =0.
∴ x =-2 or x =-1.
Hence, the given equation has 4 solutions.

15.For the equation |x2|+|x|-6 = 0, the roots are

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A. coincident
B. real with sum zero
C. real with sum 1
D. none of these

Answer: Option B.
Explanation:
|x2| =x2
∴ Given equation is
x2+|x|-6 =0.

Case I: When x≥0
In this case, |x| =x.
So the equation is
x2+x-6 = 0.
∴ (x+3)(x-2) =0.
∴ x =-3 or x =2.
∴ x = 2.

Case II: When x<0
In this case, |x| =-x.
So the equation is
x2-x-6 = 0.
∴ (x-3)(x+2) =0.
∴ x =3 or x =-2.
∴ x =-2.

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16 .If α, β are the roots of the equation x2-3x+k = 0, then the value of k for which α =
2β is

A. 2
B. -3
C. 3
D. 1

Answer: Option A.
Explanation:
α+ β =3, α = 2β

⇒ 2β =3,
⇒ beta; =1.
∴ x =1 is a root of x2-3x+k =0.
∴ 1-3+k =0
⇒ k =2

17. If the sum of the squares of the roots of the equation x2+2x-p = 0 is 10, then the
value of p is :

A. -3
B. 3
C. 6
D. -6

Answer: Option B.
Explanation:
α+ β =-2, α β = -p.
Also, α2+ β2 =10.

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⇒ (α+ β)2-2α β =10.
⇒ 4+2p =10.
⇒ p =3.
18.The sum of 2 numbers is 9 and the sum of their squares is 41. The numbers are:

A. 4, 5
B. 1, 8
C. 3, 6
D. 2, 7

Answer: Option A.
Explanation:
α+ β =9, α2+ β2 =41.
∴ α2+ β2 =(α+ β)2-2α β
⇒ 41 =81-2α β
⇒ α β =20.

(α- β)2 =(α+ β)2-4α β
⇒ (81-80) =1.
⇒ α- β =1.

Solving α+ β =9 ,α- β =1, we get α =5 and β =4.

19. The roots of ax2+bx+c = 0 will be reciprocal to each other , if

A. a =1/c
B. a =c
C. b =ac
D. a =b

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Answer: Option B.
Explanation:
Let the roots of ax2+bx+c =0 be α and β.
Then, α β =c/a.
When β =1/α, we have α β =1.
∴ c/a =1.
⇒ c =a.
20.The value of k for the which roots α, β of the equation x2-6x+k = 0 satisfy the
relation 3α+2β =20, is

A. -8
B. 8
C. 16
D. -16

Answer: Option D.
Explanation:
α+ β =6.
3α+ 2β =20
⇒ α+ 2*6 =20.
⇒ α =8.

Now, α =8 satisfies x2-6x+k =0.
∴ 64-6*8+k =0 ∴ k =-16.

21.If the roots of the equation x2-px+q = 0 differ by unity, then

A. p2 = 4q+1
B. p2 = 4q-1

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C. q2 = 4p+1
D. q2 = 4p-1

Answer: Option A.
Explanation:
α+ β =p, α β = q and α- β =1.

(α+ β)2-(α- β)2 = 4 α β
⇒ p2-1 =4q.
⇒ p2 =4q+1.

22. For what value of p, the difference between the roots of the equation x2-px+8 = 0
is 2?

A. ±2
B. ±4
C. ±6
D. ±8=p, α β = 8 and α- β =2.

Answer: C
Explanation:
(α+ β)2-(α- β)2 = 4 α β
⇒ p2-4 =32.
⇒ p2 =36. i.e p =±6

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CHAPTER 05

 Difference between a Conditional Equation and an Identity Equation.
Conditional is true for some specific values while Identity is a universal
Equation.

 In a universal Equation the sign of Equality used is " ≡ ".
 Difference between a Proper and Improper Rational Function.
 Theorem of Equality of Polynomials.(Page:180).

Now to solve MCQ’s of Partial Fractions Quickly, you have two options.

1) Just keep one thing in mind that as we can obtain
partial fractions of a rational Function conversely we
can get original Function by adding Partial fractions.
For example if Partial fractions of a function are asked
with four options, add partial fractions in each option
to get original fraction.

2) you can solve the MCQ of Partial Fraction by following
Method:

 Put x=0 in the question and check what value is coming (Note this
value). If 0/0 form is becoming by putting x=0 then put x=1,2 or 3
etc. and note the value.

 Now put that value of x in the options of MCQ for which you have
noted the value in 1st step and evaluate. For one option, the value
will match with the value of the question you calculated. It will be
the answer.
-Proper Rational Fraction:
Let P(x)/Q(x) be a rational fraction then P(x)<Q(x) is a proper
Rational Fraction.

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-Improper Rational Fraction:
Let P(x)/Q(x) be a rational fraction then P(x)≥ Q(x) is an improper
Rational Fraction.
Note: For making improper Rational Fraction, a proper Fraction,
we divide it.
For Example, please see Textbook page# 179.(Def. of Improper
Fraction)
Case#1: Resolution of P(x)/Q(x) into partial Fractions
when Q(x) has only non-repeated factor:
The Polynomial Q(x) may be written as:
Q(x)=(x-a1)(x-a2)……(x- an) whereas a1≠ a2≠…..≠ an
Then P(x)/Q(x)= + + ⋯ +

For this case, there is a shortcut (Cover Up Method) to find the
Partial Fraction.

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See the Example#1, page#181 (Alternative Method)
Case#2: When Q(x) has repeated linear factors:
If the polynomial has a factor (x-a)2 ,x≥2 and n is a +ive integer
then P(x)/Q(x) may be written as the following identity,

:. P(x)/Q(x)= + + ⋯ +

For Example, please check the Textbook page#184,Example#1
Case#3: When Q(x) contains non-repeated irreducible Quadratic
Factor:
If the polynomial Q(x) contains non-repeated irreducible
quadratic factor then P(x)/Q(x) may be written as identity having

partial fractions of the form , where A and B are the

numbers to be found.

See the example on page#186 (Text Book), Example#1.
Case#4: When Q(x)= has repeated irreducible Quadratic Factors:

If The polynomial Q(x) contains repeated irreducible quadratic
factors (ax2+bx+c)n, n≥2 and n is +ive integer then P(x)/Q(x) may
be written as the identity.

(See the example on Page#188, Example#1)

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CHAPTER 6
(Series and Sequence)

Sequence of a Function: A sequence of a function whose domain is the
subset of set of natural numbers and range is subset of real or complex
numbers.
Real Sequence: A sequence whose range (terms) is a subset of real
numbers is called real sequence.
Series: A series is obtained by adding or subtracting the terms of a
sequence.
Progression: If the terms of a sequence follow certain pattern then the
sequence is called a progression.
Types of Progression:
1) Arithmetic progression (A.P)
2) Geometric progression (G.P)
3) Harmonic progression (H.P)
1) Arithmetic Progression:
A sequence whose terms increase or decrease by a fixed number is
called Arithmetic Progression. The fixed number is called common
difference of the A.P.
Note:
1) Common difference (d) can be a +ive or –ive number.
2) Total terms (n) can never be a negative number.
-Important Results:
1) If an A.P has n terms then nth term is called last term or limiting
value of A.P and it is denoted by l or an. It is given by an=a1+(n-1)d
For Example: 1,3,5,7,9,…. Then find a11

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Solution:

Using the formula:-

an=a1+(n-1)d , as d=3-1=2

a11=1+(11-1)(2)

a11=1+20
a11=21 (Answer)
2) Three numbers a,b,c are in A.P if and only if b-a=c-b i.e a+c=2b
3) If a,b,c are in A.P then Common Difference (d) is given by:
d=c-a/2
For Example: If 2,4,6,8 then find the d?
Solution:
Using the formula:-
d= 6-2/2  d=2 (Answer)
4) If am and an are two terms of an A.P whereas am is bigger term and an
is smaller term. Also m is no. of bigger term and n is no. of smaller

term. Then d=

For Example: If a23=76 and a19=20 then find the d?
Solution:
Using the formula:
d= 76-20/23-19
d=56/4  d=14 (Answer)
5) If 1/a,1/b, 1/c are in A.P then b=2ac/a+c
6) If 1/a, 1/b, 1/c are in A.P then d=a-c/2ac
7) If a is the first term and d be the common difference of an A.P
having m terms then nth term from the end is (m-n+1)th term from the
beginning. Thus nth term from the end is given by an=a1+(m-n)d
whereas m is total terms and n is specific term from the end.
For Example: If 2,4,……,10 then find the 4th term from the end?
Solution:
Using the formula:
a4=2+(5-4)(2)

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a4=2+2 a4=4 (Answer) +1
8) For total no. of terms of A.P, we use the formula n=

For Example: If 2,4,………,100 then find the total no. of terms?
Solution:

Using the formula:
n= 100-2/2 +1
n= 49+1 n=50 (Answers)
9) Any three numbers in an A.P can be taken as a-d,a,a+d and any four
numbers in A.P can be a-3d,a-d,a+d, a+3d, any five numbers in A.P can

be taken as a-2d,a-d,a,a+d,a+2d.
For Example: The sum of three numbers in A.P is 15 and their product is
80, the largest number is? A) 2 B) 5 C) 8 (Correct) D) 11

Solution: As we know that three numbers are a-d,a,a+d
So that Sum of three number will be a-d+a+a+d=15 a=5
and the product will be (a-d)(a)(a+d)=80 a(a2-d2)=80 (5)(25-d2)=80
125-5d2=80  -5d2=80-125  -5d2= -45 d=3
Now; a-d,a,a+d=15  5-3,5,5+3 so that Largest number is 8 (Answer)

10) Middle term of three consecutive terms of A.P. is A.M. between the
extreme terms.

-Sum of First n terms of an A.P:
The sum of first n terms of an A.P with first term and d is given by:
Sn=n/2 [2a+(n-1)d]
For Example: If 2,4,6,8,10 are in A.P then find the Sum of these terms?
Solution:
Using the formula:

As n=5 , d=2 ,a1=2
S5=5/2 [2(2)+(5-1)(2)]
S5=5/2 [4+8]
S5=30 (Answer)
Note:

1) If Sn is the sum of first n terms of an A.P whose first term is a and last
term is l or an then;

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Sn= n/2 [a+l] Or Sn=n/2 [a+an]
2) If Common difference is d, number of terms are n and the last terms
is l are given then; Sn=n/2 [2l- (n-1)d]
3) If Sn=an+Sn-1
-Important Note:
1) If nth term of a sequence is a linear expression in n then sequence is
an A.P
For Example: If 1+3+5+…….+(2n-1) then find nature of progression?
Solution: As nth term is linear expression so that it is an A.P
2) If the sum of the first n term of a sequence is a quadratic expression
in n then the sequence is an A.P.
For Example: The sum of n terms of series is n(5n-1), find the nature of
the progression?
Solution:
As the sum of n term is quadratic expression so that it is an A.P
3) When the middle term of A.P is given and no. of total terms is an odd
number then sum is given by
Sum =Middle term× Total terms
For Example:
If seventh term of A.P is 10 find the sum of first thirteen terms.
Solution:
Using the Formula:
Sum=10*13
Sum=130 (Answer)
4) When the no. of total terms is an even number and the middle terms
are two then Sum is given by:
Sum = Total Terms * (middle terms/2)
For Example: If 23+24+25+26+27+28+29+30
Solution:
Here in this case Total terms are 8 and the middle terms are 26 and 27.
Sum= 8*(26+27/2)
Sum=8*(26.5)
Sum=212 (Answer)

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Note: These both tricks (3,4) are only for sum of consecutive terms.
5) Shortcut process for finding the sum of an A.P:

Once we get the corresponding terms for any A.P, we cane easily find
the sum of an A.P by using the property of averages.

Sum= Number of terms × A.M of that A.P
For Example:
For an A.P, 2,6,10,14,18,22. Sum=?

Solution:
As A.M = 22+2/2 A.M= 12
So that Sum = 6×12=72 (Answer)

6) When d=0 then Sn=a+a+a+……… n terms = na
For Example:- If 2+2+2+………..78 terms is an A.P then find the sum and

common difference?
Solution: S78=2+2+2+…….= 78(2)= 156 (Answer)
-Trick For Guessing the answer of the sum of A.P when it is given in n-

form in the options:
For Quick Guessing, use the trick:
Firstly, Put n=1 in the given options and see the sum of 1st term.

Then, Put n=2 in the given options and see the sum of first 2 terms.
Then, Put n=3 in the given options and see the sum of first 3 terms.

For Example:
If 1,3,5,7,……..+(2n-1) is in A.P then sum of this A.P is:
A) n2 (Correct) B) n(n+1) C) 2n+1 D) None of these

Solution:
As Correct Answer is Option So we use this trick on it. But when we are

in Exams we check it for each option.
Using the trick:
Put n=1 in given options i.e
Option. A: (1)2=1 and check the sum of 1st term that is 1
Put n=2 in given Option i.e
Option A: (2)2=4 and check the sum of First 2 terms that is 4

and so on.
Note: We don’t need to check it for more than first 2 values of n

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because if it’s satisfied with the given option then that’s our Correct
Answer. (Checking for more than first 2 values (i.e n=1,2) is just waste
of time).
-Important Note:
(This trick is also valid for sum of n-term of G.P)
- Note:
Sum of infinite series of an A.P does not exist.
-Properties of A.P:
1) If a1,a2,a3,……an are in A.P then;
a) a1+k,a2+k,a3+k,……… an+k are in A.P
b) a1-k,a2-k,a3-k,………….an-k are in A.P
c) a1.k, a2.k, a3.k………..an.k are in A.P
d) a1/k, a2/k,a3/k,………an/k are in A.P , k≠0
2) If a1,a2,a3,……an and b1,b2,b3,……bn are in A.P then the basic
mathematical Operations can be operated on it.
3) If a1,a2,a3,……an are in A.P then;
a) a1+an=a2+an-1=a3+an-2
Note: In an A.P. of finitely many terms, sum of terms equidistant from
the beginning and end is constant equal to the sum of the first and last
terms and so on.
For Example: If a1+a5+a11+a15+a20+a24=225 are in A.P then find the Sum
of first 24 terms?
Solution: Using the property:
As, If a1+a5+a11+a15+a20+a24=225
So; a1+a24=a5+a20=a11+a15
Hence a1+a24+a1+a24+a1+a24=225
3(a1+a24)=225
a1+a24=75
Then; we know that Sn=n/2 [a1+an]
So that S24=24/2 [75]
S24=12[75]
S24= 900 (Answer)
b) an=(ar-k + ar+k)/2 , 0≤k≤n-r

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For Example: 2,4,6,8,10,12 then find the third term?
a3=(a3-2 +a3+2)/2
a3= a1+a5/2
a3=2+10/2
a3=6 (Answer)
-Important Result:
1) If a2,b2,c2 are in A.P then , , are in A.P.

2) If l,m,n be the pth,qth and rth terms of an A.P then
l(q-r)+m(r-p)+n(p-q)=0
and p(m-n)+q(n-l)+r(l-m)=0
-Inserting Single Arithmetic Mean (A.M):

A.M=

For Example: If 2,10,b then find the value of b:
Solution:-
As we know that:

A.M=

10=

20=2+b
20-2=b
So b=18 (Answer)
-Inserting n-Arithmetic Means b/w two given numbers.
Am= +

Whereas;
m=no. of specific A.M
n=no. of A.M b/w a and b
 Find the problem related to it yourself 
-Note:

General Formula of nth A.M b/w a & b is also given by; An=

-Important Notes:
1) The sum of n arithmetic means b/w two given numbers is n times

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the single A.M b/w a and b. i.e.
A1+A2+……+An=n(A.M) i.e. Sum of n-A.M= n(a+b/2)
For example: If 2,A1,A2,10 then find the sum of Arithmetic mean b/w 2
and 10?
Solution:
As we know that:
A1+A2+…..An= n(a+b/2)
So that; A1+A2= 2(2+10/2)
A1+A2=2(6)
A1+A2=12 (Answer)
2) Sum of Alternating Series:
a) Sum of Series a-a+a-a+a... having even number of terms is equal to 0.
b) Sum of series a-a+a-a+a… having odd number of term is equal to a.
-Word Problems on A.P:-
-Entry Test MCQ:
A clock strikes twice when its hour hand is at one, four time when it is
at 2 and on . How many times does the clock strike in 12 hours?
Solution:
Using the condition:-
When clock strikes 1 it is twice i.e 1*2=2
When clock strikes 2 it is four times to prior i.e 2*2=4
When clock strikes 3 it is six times to prior i.e 3*2=6
and so on…
From this condition, we conclude that it’s an A.P with common
difference=2
2+4+6+…..+24
Using the Sum of first n terms:
Sn=n/2 [a1+an] S12= 12/2[2+24] S12=6[26] S12=156 (Answer)
Note: (Every Clock Related Problems are in A.P so we use the sum of n-
terms formula in A.P.
2- Geometric Progression:

A sequence of non-zero numbers in which every term except the first
one bears a constant ratio with its preceding term is called a geometric

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progression.
The constant ratio is called the common ratio and it is denoted by r.
Note:
1)r= ,∀ , > 1

2) 0 cannot be common ratio of G.P
3) No term of G.P can be 0.
Important Results:
1) If a G.P has n-terms then nth term is called last term or limiting term
and it is denoted by l or an that is given by l=arn-1
For Example: If 2,4,8,16,…a8 then find the a8?
Solution:
As we know that:
an=arn-1
As a8=ar8-1
a8=ar7
Here r= 4/2 r=2
So that; a8=2(2)7
a8=2(128) a8=256 (Answer)
2) Three numbers a,b,c are in G.P if and only if b/a=c/b i.e b2=ac
3) If am and an are two terms of an G.P whereas am is bigger term and an
is smaller term. Also m is no. of bigger term and n is no. of smaller

term. Then; r=

For Example: If a2=9 and a5=243 then r=?
Solution:
Using the formula:

r=

r=(27)
r=3 (Answer)

4) If 1/a,1/b,1/c are in G.P then common ratio is r=±

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5) The nth term from the end of a G.P with last term l and common

ratio is .

6) If a is the first term and r is the common ratio of a finite G.P
consisting of m terms then the nth term from the end is given by arm-n
For Example: If 2,4,8,16,32 then find the 4th term from the last?
Solution:
Using the formula:
a4= (2).(2)5-4
a4=2(2)a4=4 (Answer)

7) For total terms of G.P , the formula is given by rn= .

For Example: If 2,4,8,16,..,1024 is in G.P then find the total number of
terms?
Solution:
As we know that;

rn= .

2n= . 2

2n= . 2
2n=(1024)
2n=210 (As the bases are same so that n=10)
n=10 (Answer)
8) Three numbers in G.P can be taken as a/r , a, ar, Four numbers in G.P
can be taken as a/r3, a/r,ar,ar3. Five numbers in G.P can be taken as
a/r3,a/r,a,ar,ar3.
For Example:
The sum of three numbers in G.P is 19 and their product is 216 then r=?
A) 1.5 (Correct) B) 0.5 C) 1.25 D) 2.25
Solution:
As we know that:
Three numbers in G.P can be taken as a/r , a, ar
So that; Sum a/r+a+ar=19 (a+ar+ar2)/r=19

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and product  a/r.a. ar=216 a3=216a=6
so that (a+ar+ar2)/r=19
Taking common; a( 1+r+r2)/r= 19
so 6(1+r+ r2)=19r
6+6r+6r2=19r
6r2+6r-19r+6=0
6r2-13r+6=0
6r2-9r-4r+6=0
3r(2r-3)-2(2r-3)=0
3r-2=0 , 2r-3=0

r=2/3 , r=3/2
So the r=3/2 is the answer in this case.

9) Sum of Infinite Geometric Series when common ratio is 1/2 then
Sum= 2×(First Term)
For Example: 1,½ ,1/4, 1/8,1/32……is in G.P, then find S∞?
Solution:-
Using the formula:
Sum =2×1 Sum=2 (Answer)

For Example: ∑ ( ) =?

Solution:

Using the Formula:
Sum=2×1/2 =1 (Answer)
-Sum of first Terms of a G.P

The Sum of first n terms of a G.P with first term and common ratio r is
given by:

= ( ),| | > 1

Or = + + + ⋯ … . . +
For Example: 2,4,……….a10 , Find the sum of first 10 terms of G.P?

Solution:-
a=2 , r=2 , n=10

= ,| | >1

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=( )

= 2046 (Answer)

Note: ) ,| | < 1
1) = (

2) = ( ) , | | < 1

3) = , | | > 1

4) = ( ) , | | > 1

-How to solve the Sum of n-terms when Summation sign comes in the
Question:
If ∑ ( )
then Using the easy method, we can solve it within seconds.
Method:
Put first value as given below in the series n=1
then put second value n=2
and put third value n=3
now take the common ratio and then apply the Sum of n-term formula.
Note: (We start putting the value of n from where it is given, in this
case we just suppose to begin with n=1)
For Example:
If ∑ → ( ) =?

Solution:

Using the easy method:
Put n=1; (1/i)1=1/i=-i
Put n=2; (1/i)2=1/i2=-1
Put n=3; (1/i)3=1/i3=i
Put n=4; (1/i)4=1/i4=1

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Put n=5; (1/i)5=1/i5=i
so that r=1/i
As it can be simplified by simple method here we don’t need to use the
formula:

S5= -i-1+i+1-i
S5= -I (Answer)
-Sum of Infinite Series of G.P:
The sum of an infinite G.P with first term a and common ratio r is:

= 1− ,| | < 1 −1 < < 1
For Example: If 27,9,3,1,………∞ then find the sum?
Solution:-
As we know that;

=1− ,| |< 1
So that = , | | < 1,

Now; = , | | < 1

Hence; = ( )

And this series is convergent as |r|<1
-Note:
1) An Infinite geometric Series is convergent if;
lim → exists i.e. |r|<1
(Example is solved as above)
2) An Infinite geometric Series is Divergent if |r|≥1
For Example: 1,3,9,27,………… find the nature of series?
Solution:-
As; r=3/1  r=3 and it is greater than 1 so that it’s a divergent Series.
3) An Infinite geometric Series is Oscillatory if r=-1

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For Example: If 2-2+2-2+……, find the nature of Series?
A) Oscillatory (correct) B) Divergent C) Fluctuating D) Convergent
Solution:-
As we know that r=-2/2 =-1 so it’s an oscillatory series.
4) An Infinite geometric Series is fluctuating if r does not remain
constant.
For Example:-
The series 1/3.4 + 1/4.5 +.... 1/(n+2)(n+3)+.....is

(a) Divergent (b) Convergent (c) Fluctuating (Correct) (d) N.O.T
Solution:-
As it is; 1/12+1/20+1/30 +1/42+……… +1/(n+2)(n+3) +…..
r= (1/20)/(1/12)=3/5
r= 1/30/1/20= 2/3
…. And so on.
From this, we conclude that r doesn’t remains same, therefore it’s a
fluctuating series..
Note: S∞ does not exist if |r|≥1
-Ratio Test For Determining the nature of the Series:

1) If Sn = u1 + u2 + u3 + ….. + un converges as n∞ lim → <1

2) If Sn = u1 + u2 + u3 + ….. + un diverges as n∞ lim → >1

For Example: ∑ ( ) Then find the nature of Series?

Solution:

Using the Exponent Rule:

:.lim → ()
()

: . lim ( )

→ ()

:.lim → ( )

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: . lim ( )

→

:.limx∞|(-1/5)n| as |-a|=a
So applying the limit it is (1/5) which is less than 1 hence it is
convergent series.
Note: (This ratio test is valid when nth term is also given in the infinite
geometric series)
-How to solve the Question when Summation sign comes in the
Question sum of Infinite geometric Series.
If ∑ ( )
Method:-
Firstly, Put n=1
secondly, put n=2
and then put n=3
and later on take the common ratio and use the Sum of infinite
Geometric Series formula.
For Example: ∑ (− ) =?

Solution:-
Using the easy method as described earlier:-
:.∑ (− ) = (− ) + (− ) + (− ) + ⋯

As r=9/7 /
/
So that; S∞= =9/40 (Answer)

Note: (We start putting the value of n from where it is given, in this

case we just suppose to begin with n=1)

MCQ:-

Here is most frequently asked MCQ in the entry tests but this method

sometimes be not in consideration.
If y=1+2x+4x2+8x3+……. Then find x=?

Solution:

Use the Sum of infinite series formula on Right Hand side:-

As r=2x

So; y= 1/1-2x

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Now; y(1-2x)= 1
y-2xy=1
-2xy=1-y
Taking common Both sides.
-2xy= -1(-1+y)
2xy=y-1
x=y-1/2y (Answer)
-WORDS PROBLEMS ON G.P:
-Tricky MCQs:
An Object is dropped from a building having “h” vertically downward ,
after one drop it jumps to height ( ) of height (initial height),find the

distance before it comes to the rest?

( ()

Trick: Distance= ( )

For Example: What distance will a ball travel before coming to rest if it
dropped from a height of 75m and after each fall it rebounds 2/5 th of
the distance it fell
Solution:
Using the trick:
As a=2 , b=5 so that;

( ()

Distance= ( )

Distance=175m (Answer)
-Properties Of G.P:
1) If a1,a2,……are in G.P then;
a) a1 k,a2 k,a3k ,…… are also in G.P
b) a1/k,a2/k,a3/k,……. are also in G.P , k≠0
c) 1/a1 ,1/a2 ,1/a3 ,….. are also in G.P
d) a1k, a2k,…… are also in G.P
2) If a1,a2,a3,….. and b1,b2,b3,….are in G.P then;
a) a1b1,a2b2,…… are also in G.P.
b) a1/b1,a2/b2 ,…….also in G.P.

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3) If a1,a2,a3,……an then ;
a) a1an=a2an-1=……..
Note: In a G.P. of finitely many terms, the product of terms equidistant
from the beginning and end
is constant equal to the sum of the first and last terms.
-Inserting Single Geometric Mean:
G.M= ±√ab
For Example: If 3,G,27, then find G?
Solution:-
As we know that;
G= ±√3.27
G=±√81
G=±9 (Answer)
-Note: Here we take the +9 as the G.M as all the terms are positive
-Inserting n-Geometric Means b/w two given Numbers:

:. =

Whereas n=no. of G.M b/w a and b and m=specific G.M which is to
find.
 Find the problem related to it yourself. 
-Note:-

The general formula of nth G.M b/w a & b is also given by Gn=

-Important Results:
1) The Product of n geometric means b/w two given numbers is nth
power of the single G.M b/w them i.e. If a and b are two given numbers
and G1,G2,……Gn are n geometric b/w them then
G1.G2.……Gn = (√ ) or an (b/a)n/2
For Example: If 2,G1,G2,G3,32 then find the Product of G.M?
Solution:
As we know that:-
G1.G2.G3 = (√2.32)
G1.G2.G3 = (8)

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G1.G2.G3 =512 (Answer)

2) If A and G are respectively arithmetic and geometric means b/w two

positive numbers a and b then:

a) A>G

b) The quadratic equation having a and b as its roots is x2-2Ax+G2=0.

For Example: Find the quadratic equation in such the arithmetic mean

of its roots is 4 and its geometric mean is 9?

Solution:-

Using the formula:
As x2-2Ax+G2=0

So x2-2(4)x+(9)2=0
x2-8x+81=0 (Answer)

c) The two positive numbers are ± √ −

For Example: If arithmetic mean and geometric mean are 5 and 4

respectively then find two numbers on extreme of these means?

Solution:-

As we know that:

The two positive numbers are 5 ± √5 − 4

then; two positive numbers are 8 and 2 (Answer).

Entry Test MCQ

If 0.2+0.02+0.002+0.0002+……..is in G.P Find the sum of Infinite Series?

Solution: First of all, simply it.

Take common 2 for all the terms:

2( 0.1+0.01+0.001+0.0001+…….)

Now in the brackets find the common ratio i.e 0.01/0.1= 0.1

Now Again multiply 2 in the brackets and solve it for Sum of series.

As =

So; = . = . = (Answer)
.
.

-Arithmetic-o-Geometric Series (A.G.P.)

This is combination of Arithmetic and Geometric Series.

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If P1, P2, P3 ... be an A.P. and a1, a2, a3 ... be a G.P. then p1q1, p2q2,p3q3, ...
is said to be an arithmetic-o-geometric progression. A general A.G.P. is
a,(a+d)r,(a+2d)r2,(a+3d)r3,...
-Nth term of an A.G.P:
The nth term of an A.G.P is given by:-
Tn={a+(n-1)d}.rn-1
Note:- A sequence is both an A.P. and a G.P. iff it is a constant
sequence.
3) Harmonic Progression (H.P):
A sequence of numbers reciprocal of whose terms form an A.P. is
called harmonic Progression.
For Example: ¼,1/7,1/10 …. Is a H.P since 4,7,10,….is an A.P of nth
term= 1/nth term of the corresponding A.P = 1/a1+(n-1)d
-Note:
1) No term of H.P can be zero.
2) Three number a, b, c are in H.P if and only if b=2ac/a+c .
3) There is no general formula of finding the sum of n terms of H.P.
-Inserting Single Harmonic Mean:-

H.M=

For Example: If 2, H, 10 then find H.M?
Solution:
Using the formula:
H.M= ( )( )

H.M=

H.M= (Answer)

-Inserting n-Harmonic Mean b/w a and b:
Hn= ( )

For Example: 2, H1,H2, 10 then find H.M?
Solution:-
Using the formula;

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Hn= ( )
()

Hn=

Hn= (Answer)

-Relation b/w A.M, G.M and H.M:

If A,G and H be the harmonic ,geometric and harmonic means b/w a
and b , then;
1) G2=A.H

For Example: If G and A are 6 and 4 respectively then find the value of
H?
Solution:

Using the formula:
G2=A.H
(6)2=4.H
36/4=H H=9 (Answer)
2) For a and b distinct positive real numbers.

A≥G≥H, where G>0
3) For a and b distinct negative real numbers.
A≤G≤H , where G<0

4) A,G,H are in G.P
-Important Result:

1) Reciprocal of the term of G.P forms a G.P
2) Reciprocal of the term of A.P may or may not form a H.P
3) Reciprocal of the term of H.P forms a A.P

4) If a,b,c are in G.P then x,y,z are in H.P
5) If a,b,c are in G.P then 1/x,1/y,1/z are in A.P
6) If b=c then a2,b2,c2 are in G.P.

7) The number , , are in G.P.

8) The reciprocal of whose terms form again same type of sequence is
H.P.
9) If a,b,c are in A.P then 1/a,1/b,1/c are in H.P.
10) If the first term of an infinite geometric series is equal to twice of

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the sum of all the terms of that follows it, then the value of r is 1/3
11) If a, b, c form a G.P with common ratio r (0<r<1).If a,2b,3c form an
A.P then r= 1/3
12) Three numbers of G.P. If we double the middle number we get an
A.P the common ratio of G.P is 2±√3 .
13) Non zero terms are in A.P, G.P and H.P.
14) If a,b,c is in A.P then b is A.M.
15) If a,b,c is in G.P then b is G.M.
16) If a,b,c is in H.P then b is H.M.
17) Every term of G.P is the logarithm of each term of A.P.
18) Every term of A.P is the anti logarithm of term of G.P.
19) If a,b,c are in H.P then bc, ca, ab are in A.P.
20)If a,b,c are in A.P then 1/1-a , 1/1-b , 1/1-c are in H.P.
-Some Important Points:

If the expression is:

1) A.M when n=1
2) G.M when n=1/2
3) H.M when n=0

If the expression is:

1) H.M when n=-1

2) A.M when n=0

3) G.M when n=-1/2

-Important Results:
If α and β be roots of the quadratic equation x2-2Ax+G2=0 then;

1) A.M=

2) G.M=

3) H.M=

For Example: Form an quadratic equation if A.M is p and G.M is q and
also find the H.M?
Solution:

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As we know that:-

A.M=

α+β=2p----- (1)
As we also know that:-

G.M=
αβ=q2------(2)
As x2-Sx+P=0
So that x2-2px+q2=0
For H.M, we know that:

H.M=

From (1) and (2)
H.M= ( )

H.M= (Answer)

Note:
Let a and b are two numbers and if a=b then;
G.M=H.M=A.M
-Some Derived Results:
When Quadratic equation is ax2+bx+c=0 then;

1) A.M=

2) H.M=

3) G.M=

For Example: If x2-10x+5=0 then find H.M ,A.M and G.M?
Solution:-
As we know that:

A.M=

A.M= ( ) ) A.M=5 (Answer)
(

As we also know that:-

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H.M=( ( )) H.M=1 (Answer)
As we also know that:-

G.M=

G.M=

G.M=√5 (Answer)
-Sum of Some Special Sequences:-
1) ∑ [ − ( − 1) ] =
2) ∑ 1 =
3) The Sum of first n natural number is given by:-
:.∑ = 1 + 2 + 3 + ⋯ . . + = ( )

4) The sum of first n even natural number is given by n(n+1)
5) The sum of first n odd natural number is given by n2.
6) The sum of squares of first n natural numbers is given by:-
:.∑ = 1 + 2 + 3 + ⋯ + = ( )( )

7) The sum of squares of first n odd natural numbers is given by

()

8) The sum of squares of first n even natural number is given by
( )(
)

9) The sum of cube of first n natural number is given by

:.∑ = 1 + 2 + 3 + ⋯ + = ( )

10) The sum of cube of first n even natural number is given by 2n2(n+1)2
11) The sum of cube of first n odd natural number is given by n2(2n2-1)

12) 1+5+9+…….+(4n-3)=n(2n-1)
13) 1+4+7+…..+(3n-2)= ( )

14) 2+6+18+…..+2×3n-1=3n-1 _( )
15) 1+2+4+…….+2n-12n-1
16) 1×3+2×5+3×7+…….+n×(2n+1)= (

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17) 1×2+2×3+……………+n×(n+1)= ( )( ) )
18) 1×2+3×4+…………….+(2n-1)(2n)= ( )( )
19) 12-22+32-42+………….+(-1)n-1(n)2=( ) . (

20) × + × + × + ⋯ + ( ) = 1 −
(Find yourself the examples of these points and apply on it)
Note: These formulas are only valid for A.P.
Note:
1) nth term of b+bb+bbb+…(form) is
an= (10 − 1),whereas b∈{1,2,…..9}

For Example:
Find third 5th term of 3+33 ?
Solution:
Using the trick:
a3= (10 − 1)

a3= (1000 − 1),

a3= (999)=333 (Answer)

2) nth term of 0.b+0.bb+0.bbb+… (form) is
an= (1 − 10 ),whereas b∈{1,2,…..9}
For Example: Find the 4th term of 0.2+0.22+..?
Solution:
Using the trick:
a4= (1 − 10 )

a4= 1 −

a4=

a4=0.2222 (Answer)
-Rules to solve decimal fraction (forming a common/vulgar Fraction)
Here are the key points of working out Vulgar fraction within seconds.

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The sign of repeating is dot over the digit.
Case-I (When all Digits in Decimal part are repeating)
In this case , in the denominator of vulgar Fraction , the number of
nines is equal to number of repeating digits and numerator is actually
the complete given number without decimal minus the number
before decimal.
Examples:-
1) Vulgar Fraction of 2. ̇ ̇ ̇ will be : 2342-2/999 = 2340/999.
2) 13. ̇ ̇ ̇ ̇ will be : 134235-13/9999 = 134222/9999.
3) 0. ̇ ̇ ̇ will be 0271-0/9999 = 271/999.
Case-II (When all digits are not repeating)
In this case, in the denominator of vulgar fraction, the number of
nines is equal to the number of repeating digits and after nines we put
zeros and the number of zeros is equal to the number of non-
repeating digits in decimal part. The numerator is the whole given
number without decimal minus the number before repeating digits.
Examples:

1) Vulgar fraction of . ̇ ̇ ̇ will be 21341-21/9990 21320/9990 (Answer)
2) Vulgar fraction of 0.02 ̇ will be 0021-002/900 19/900 (Answer)

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CHAPTER#7 (COMBINATION, PERMUTATION
AND PROBABILITY

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